Answer:
Option D
Explanation:
Volume of big drop = n(Volume of small drop)
\Rightarrow \frac{4}{3}\pi R^{3}=n.\frac{4}{3}\pi r^{3}
where, R= radius of big water drop
and r = radius of small water drop
\Rightarrow R^{3} =n r^{3}
\Rightarrow R= n^{1/3}.r ........(i)
surface energy of n drops,
E_{2}= n \times 4 \pi r^{2}\times T
Surface energy of big drop ,
E_{1}= 4 \pi R^{2}T
\therefore \frac{E_{2}}{E_{1}}= \frac{nr^{2}}{R^{2}}= \frac{nr^{2}}{(n^{1/3}.r)^{2}}
= \frac{nr^{2}}{n^{2/3}.r^{2}}= n^{1/3} [\because from eq.(i)]
E_{2}:E_{1}=\sqrt[3]{n}:1